Question

The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110.
a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?
b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?
c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?
d. What is the cost for the 3% highest domestic airfares?

Answer 1

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(domestic airfare is $550 or more)

P(x > 550)

$P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)$

$= 1 - P(z \leq 1.5)$

Calculation the value from standard normal z table, we have,  

$P(x > 550) = 1 - 0.933 = 0.067= 6.7\%$

b) P(domestic airfare is $250 or less)

$P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)$

Calculating the value from the standard normal table we have,

$P( x \leq 250) = 0.111 = 11.1\%$

c))P(domestic airfare is between $300 and $500)

$P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%$

$P(300 \leq x \leq 500) = 61.2\%$

d) P(X=x) = 0.03

We have to find the value of x such that the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 385}{110})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 385}{110})=0.03$

$=P( z \leq \displaystyle\frac{x - 385}{110})=0.997$

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 385}{110} = 2.748\\x = 687.28$

Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.